Node Voltage Problems in Circuit Analysis - Electrical Engineering Node Voltage Analysis Problem
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- Опубліковано 14 жов 2024
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Learn what the node voltage method is in circuit theory and how to use it to solve circuits. First, we will describe what nodal analysis is in electrical engineering circuit analysis. Next, we will present a problem and write down the node voltage equations necessary for the solution. Finally, we will solve the node equations and discuss the results..
You state the tiny details NO ONE MENTIONS! I freaking appreciate your videos! I am liking and subscribing!
The way you keep repeating these little details over and over throughout the video makes them stick really deep in my mind! Awesome tutor and a Brilliant tutorial :)
For real!! Not only once, that also happened to be said when one sneezed so one misses it!!!
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This has been an excellent tutorial about applying node voltage method into solving circuit analysis problem. The presentation of the material was professional and the author did a very good job of explaining and justifying what he was doing every step of the way. Although the circuit was pretty simple, the fact that the author labelled everything in the circuit make it clear for students to understand what he is talking about. I really appreciate that the author uses different pen color on the whiteboard. It make it so much easier for student to understand the equations he wrote down. Thank you so much!
I love you
Love who? Stop cheap humor on educational video.
If your an engineering student you know he's board is freaking awesome!
Lol thanks, yes I am proud of it :)
That's right.. He's the best for me
One of the best explanations of circuit analysis ive probably ever come across. thank you so much.
He is such an awesome instructor from bottom to top of learning the subject matter including the techical analysiz. He is soooo good unbelievably extraordinary..
I did my study in electrical course but none our teachers can be like him.. If we can not understand we we were asked to make research which is very difficult..
That ( SIGN + / - ) known first or all answers in the equation / s made were all wrong..
Thank you Sir Jason reminding us of that MANDATORY SIGN.
Very well put! This is the most helpful video I've found on this. The extensive explanations are just what I need. Thank you.
Fantastic tutorial! Wished my tutorials in college were this detailed 😊👍
Thank you. This was a great tutorial! My professor is awesome but there is a language barrier so I miss a lot in lecture. This will be my secondary lecture. Thanks again!
I want this professor in my online class🙂
Bcz of their method is just awesome nd board work also well explained.... 👌🙏
Finally grasping how to use node analysis to solve circuits and let me just say this video is really good at showing how it's done.
I was so confused over sign convention's. Thanks for helping me out in one time. Done so perfectly!!
this channel is sooo much underrated
Amazing amazing amazing!! Explained so properly and easily. This sir, is a talent like no other
I really appreciate it!
Small but important point. Go ahead and show the direction of your currents...
agreed
Thanks a lot I've been struggling A LOT to solve my midterm project questions because my professor didn't teach even ohm's law in online classes and said research and solve yourself... I was feeling really overwhelmed until I watched this. You are a really great teacher (:
Why will your professor give you what he didn't teach you!!
You are by far the best at teaching this
Thank you! this made me help remeber things I forgot about nodal analysis. I thought solving the final equations with a Simultaneous equations method is easier than the matrix.
Most helpful video about nodal analysis..Many many thanks.
Thank you so much!! The repetition and step by step is EVERYTHING one needs!!!!!! Very helpful!!!!
I just love the way he teaches. Really saved my arse for exams!
Awesome! Lets see if I can incorporate this into my test today!!!!
Excellent presentations. I am glad I found these videos! Great work. Thank you!!
Tip for people without access to a calculator to set up simultaneous eqn.'s: create whats called an augmented matrix by (this is very rigid and has a formal definition) lining up the coefficients and RHS's and create a matrix out of that. In this case
[ 6 -5 240]
[-30 35 -240]
We want to create a simpler but equivalent matrix byidentifying the pivot columns and creating 0's under pivots using elementary row operations (where a pivot is the leftmost entry that's nonzero in a row, a pivot column being having only 1 pivot entry in the column). In this case, all we need to do is perfom 1 row operation to get the information we need. Adding 5x row 1 row 2 we get:
[6 -5 240]
[0 10 960]
We can convert this back to our system of equations and solve for v1 and v2.
10v1 = 960 -> v1 = 96
6v1 - 480 = 240 -> 6v1 = 720 -> v1 = 120V.
Wasn't sure if people already knew how to do this but I always find this to be the most intuitive approach to solving simultaneous equations as long as the system is small.
Good video just to be clear. V1 is 119V (engineering notation) or 119.28V (scientific notation) V2 is 96. Good video over all.
This tutor is the BEST!
I solve these problems a little differently. Instead of pretending that all currents are leaving the principal node for each equation, I draw arrows on my circuit diagram as a first step indicating the directions of the branch currents. Then the voltage difference between any two nodes is always the node voltage at the tail of the arrow minus the node voltage at the head of the arrow (regardless of whether the current in question is entering or leaving a particular node). Next, I write the KCL equations as the sum of currents entering the node of interest = the sum of currents leaving the node of interest. Then I find a common denominator on the LHS and a common denominator on the RHS, and cross multiply them to eliminate the fractions. I move all variable terms to the left and all constant terms to the right, and solve the system. Always works, and I don't have to worry about sign errors.
The easier way I found was to write your current analysis equations as flow in and flow out. you end up with the same equation when you re-arrange. Great video :D
You have the best way of explaining. Thank you.
This guy have saved my bacon. Cal 1-3, physics, and diff equ. Without him, I would be lost in all of my classes
Thank you so much professor. Your explanation very clear and now i know how to write the equation and solving using gaussian elimination method. I owe you big one!
Great tutorial!! Your explanations are beginner-friendly.Thanks alot never knew circuits could stop seeming complex to me.
So happy you liked it!
I didn't actually understand this in our class an hour ago. Thanks for the video!
Great job! This will definitely help me pass my exam today! Thanks!!!
That multiplication with the same number on RHS and LHS could go a bit lengthy, there is an easy way though. we can write the equation 1 as: (1/8+1/80+120)*V2 - V1/8 + 1 = 0 . i just took all the v1 and v2 terms common and solve . this could actually help you in eliminating the big numbers (like 240) in the above video.
explained very clearly, i solved final part using Cramers 2 detrminate rulls.
Oh my gosh, thank you. The wires with no resistors were absolutely killing me.
I knew I would find some understandable explanation on this subject! My professor did not help me at all! Thanks for the video. It saved a few nights up, and some cups of coffee :P. Good job!
Very great explanation! although camera movement a bit distracting but the idea was very well explained. Thanks!
Really you help me a lot i have just watched your starting and gona download your video for reference, Thanks for uploading.
This guy is a KING; he just doesn't know it yet 😍😍
THANKYOU VERY MUCH ! =D it took me a long time to find out how to solve the voltages after using system of equations until now :)
couldn't have explained it better, thanks and gig em
Thank u sir very clear n easy to understand God bless u sir
Hello! Ur videos have been so helpful :) my grades went from C to A. Could u please upload more videos on circuit analysis ??
you're a rockstar dude!
The last Matrix method in which He used to solve for V1 and V2, Is it necessary to use that method while solving in Node Voltage Method or Is it just an optional??????????
No it is not necessary. You can solve it by other algebraic methods instead of matrices.
Thanks bro....all the tiny bits of info were really helpful
i have 2 questions
1-do u have a video helping with resistors in parallel or series and helping to understand them.
2-do u have any videos with matrices?
Did you use the inverse matrix method for a particular reason? I set up the equations as an augm. matrix, because I noticed all the numbers were multiples of 5, which made the row operations quite simple..
You are awesum sir , loved your method of teaching .
Thank you very much, I was struggling with the sign convention but you really cleared it up!
Plz make a video with all methods,should contain short explanation and main problems,like if everyone watch that could solve any problems topics about methods problems like mesh Norton and so on in one video)
This man is a Legend.
very informative for node analysis. thanks...
Thank you very much for this video. It is super clear!
This is a great lecture though I may add that you should label the assumed currents caused I sometimes get lost what operations should I use. Anyways, Thank you
Nice explanation👍
Great video and example.
U can just apply supermesh method or source transformation by nodal take just one node at 80ohm and apply KCL.
Hello sir i appreciate your hard work and help towards the students. I've got a small question. In the equation one you have used (v1-v2)/8 for the current across 8 ohm resistor.but in the equation 2 you have used (v2-v1)/8 for the current across 8 ohm resistor. shouldn't it be the (v1-v2)/8 for every equation?
+Rumesh NIlanka In the first equation as we are taking the operational node as node 1, so all the currents we take are going away from the node ie. going from higher potential to the lower potential .so, node 1 is at higher potential and we take (V1(higher potential)-V2(lower potential))/8 across 8 ohm. Same is the case with node 2. in eq. 2 ,as we take all the currents going away , node 2 will be at the higher potential and node1 at the lower. hence, (V2(h.p)-V1(l.p)/8.
+abhilash vana So in node 1, we take current going right ( from v1 to v2). And in node 2, we are taking the current going left( from v2 to v1)? Am I understanding this right? If in node 2, we take current going right (towards the 120) then the equation would be -(v1-v2) /80??
+Toiletwifi open the braces . you will be getting the same but not 80 its 8 ohms resistor between those two nodes (v1 and v2)
Yeah I know it will be the same, but is my logic correct? If we put the current direction going right, at node 2 eq will be :
-(v1-v2)/(8). The negative is there because we are assuming that current going IN a node= -ve convention. And we do v1-v2 because curernt is leaving FROM v1. Am I understanding this right? and Oh yeah I meant 8, not 80*
You're a legend.
thank u that is was very helpful.... but i have one question which is,what i have to do when the voltage source is connected in series connection with a resistor??? thx again ^_^
Hi, I was a little confused, my teacher taught us that it was current in = current out of a node, which usually meant that when you picked a direction for current between two nodes, you kept that same direction each time, so it was v1-v2 for all nodes. Is this different than assuming all currents are leaving the node?
Seems really easy to mess up nodal analysis if certain things get switched around.
it works out the same if you use correct sign conventions, a-b=0 is the same as a=b, I personally think the way this guy does it is easier
ya but it dose't mean a-b=0 same as b-a=0.I also a little confused at that point
It's the same, but this guy's technique is superior in my opinion. I prefer not to add sign conventions to add confusion to the analysis. Just assume the current leaves each node and the sum of that is ZERO. Current source entering a node is (-) in the equation. Likewise, current source leaving a node is (+) in the equation.
pronob roy how can it not be? if x -y yields zero then y-x will also yield zero
it implies that y is equal to x, first understand what you're writing
fantastic walkthrough. thank you
Thank you in this lesson!
Great tutorial, as always! But lost me at the inverse matrix. ☹️
GREAT video, thanks a lot :) Everyone else I found online was monotone, slow and unclear :D
nice video but I have a question why did you write (v1-v2)/8 instead of (v2-v1)/8 aren't we assuming the current to be going out of the node and the current goes from -v to +v ?
What if I’m a particular problem, v2 is turns out negative? I would imagine voltage travels backwards from what expected, or is it a mistake?
thank you so much sir for your clear explanation of this
While using the matrix method of solving certainly works and may be easier in more complicated circuits, _Gaussian Elimination_ technique seems to be perhaps an easier way to solve the unknown voltages than using the seemingly more tedious inverse matrix method, at least on simpler circuits like that which is presented. Just my opinon.
I prefer using simultaneous equations rather than matrix but thanks man! good tutorial!
Thanks a lot it really helped me. keep it up
Thank you sir. God bless you. Amen.
Delicious knowledge that I so needed. Thank you good sir!
you are the best
I wish you teach in our college man
Thank you so much!
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you saved my life
I am definitely subscribing.
When I put carefully this problem on simulator, Both V1 and V2 have a negative sign (-120v and -96V)
I keep getting v1= 20, v2 = -24 .. I plugged the same 2 equation you got in a calculator and did it by hand and it gave me the same results.
+Aqeel Almousawi i know, i got the same too. but my V1 IS 60V. Did you figure out what went wrong?
+Antony Muchina you are doing something wrong, the way you can check is by doing the matrix by hand.... do a cross multiplication, then subtract them....
(6*35)-(-30*-5)=60 this will give you the denominator
then by doing the cross multiplication with the 240 and -240 will give you the following:
(6*-240)-(-30*240)= 5760
(-5 * -240) - (35 * 240) = -7200
then just divide them by 60
5760/60= 96
-7200/60=-120
and thos would be your values for V1 and V2
OMg you are better then my sir :)
This was very useful. Thank you.
I wonder the two on the right that are in parallel, could you combine them into their equivalent resistance and then treat v2/ 48 like you did v2/80 & v2/120 in eq 2? Since they have the same voltage and all?
Went ahead and did it, actually works!
you get:
6v1-5v2-240=0
-6v1+7v2+48=0
6's cancel and it solves straight away. Easier than having to guess where the current's going and what to do with that middle 80 ohms.
thanks ill buy your course i think :)
The explanation was soo good but V1= 2.5, V2= 9
nice video!
This is really helpful. Thanks.
very very very useful , really thank you
Very thorough. Thank you!!
Welcome!
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you can even use the calculator to directly enter the equation and you can get the voltage value
So helpful! Thank you sir!!
very useful... thank you sir
thanks, this video helps me a lot.
what happens when we have two voltage sources connected in one single node? Do you create a supernode that contains both?
hey did you use a algebraic equation?
Thanks a lot, this helped me out very well
This guy is awesome!
Great video. Thanks a lot!